Determinants

Given that a = (-1, 3, 7), b = (4,-2,-9) and c=(5,5,3), work out a x b (a cross b) and show that it is perpendicular to c.

I worked out a cross b using the determinant method, and got -13i -19j -10k, now to show its perpendicular to c I do

[(a x b) . c]

but when I do this, I get -190, which clearly isnt 0, the answer I need to show that its perpendicular, anyone help?

Reply 1

JamesF

1

I get the same answers as you. Dodgy question maybe?
I think c = (5, -5, 3)

Reply 2

Barny

OP

17

Could be dodgy, but Im not getting the right answer either way!!

Reply 3

JamesF

1

If c = (5, -5, 3),
(axb).c = 5*(-13) -5(-19) + 3(-10) = -65 + 95 - 30 = 0

Reply 4

JamesF

1

So its probably just a typo on the problem sheet.

OP

imasillynarb

Given that a = (-1, 3, 7), b = (4,-2,-9) and c=(5,5,3), work out a x b (a cross b) and show that it is perpendicular to c.

I worked out a cross b using the determinant method, and got -13i -19j -10k, now to show its perpendicular to c I do

[(a x b) . c]

but when I do this, I get -190, which clearly isnt 0, the answer I need to show that its perpendicular, anyone help?

I get a x b is (-13, 19, -10) not (-13, -19, -10) I worked that out using the regular formula though...
to use the determinant method:-

i j k
-1 3 7
4 -2 -9
i component :- (3x-9 - 7x-2) = -27 + 17 = -13
j component :- (7x4 - -9x-1)= -(28-9) = 19
k component :-(-1x-2 - 3x4) = (2-12) = -10
=>(-13,19,10)

[When you are doing the j component you just (imaginarily) shift the k column to the left of the i column to give:-

k i j
7 -1 3
-9 4 -2

So then you will have 7x4 - (-1 x -9) = 19]

Google search gave me this website:-
http://www.ucl.ac.uk/Mathematics/geomath/level2/mat/mat121.html
which does the same thing
Katie

Determinants

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